Question

In an electrical cable, there is a single copper wire of radius $6mm$. Its resistance is $10\Omega$. The single wire in the cable is replaced by five different insulated copper wires with the radius of $2mm$ for each wire. The new resistance of the cable is:

• A. $50\Omega$
• B. $18\Omega$
• C. $90\Omega$
• D. $12\Omega$

Answer 1

The correct option is B $18\Omega$

The resistance of the original single copper wire,

$R_1=\frac{\rho l}{A_1}=\frac{\rho l}{\pi r_1^2}$, where $r_1=6mm$ and $R_1=10\Omega$
The resistance of each new copper wire,

$R_2=\frac{\rho l}{A_2}=\frac{\rho l}{\pi r_2^2}$, where $r_2=2mm$

The effective area of 5 such new copper wires, say $A_3$ is$=5\times A_2$

The resistance of new copper wire consisting of 5 wires is,

$R_3=\frac{\rho l}{A_3}=\frac{\rho l}{5\pi r_2^2}$

The length $l$ and resistivity $\rho$ is same,

$\therefore \frac{R_3}{R_1}=\frac{A_1}{A_3}=\frac{\pi r_1^2}{5\pi r_2^2}$

$\Rightarrow \frac{R_3}{R_1}=\frac{(6{)}^2}{5(2{)}^2}=\frac{36}{20}=1.8$

$\Rightarrow R_3=1.8R_1=1.8\left(10\right)=18\Omega$

Option (B) is correct.